Wikipedia:Reference desk/Archives/Mathematics/2015 July 31

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July 31

Cosets and normal subgroups

A couple of months ago, an editor added a claim to normal subgroup that a subgroup N of a group G is normal if and only if the collection of left and right cosets of N coincide. (It has since been removed.) One of the two implications is obvious, but the other direction seems very dubious to me; does anyone have a simple counter-example? --128.101.152.166 (talk) 15:34, 31 July 2015 (UTC)

It's true. If N is normal then gN = Ng implis that every left coset is a right coset. Conversely, if every left cost is a right coset, then gN = Nh for some h. So, since g belongs to the coset Nh, Nh=Ng as well. Thus gN=Ng. Sławomir
Biały^

Thanks. I disagree with your adding it back to the article, however, unless you have a source that actually defines normal subgroup this way (and it is mentioned in the body somewhere). That is, I think the deduction of the equivalence is nontrivial, even though not very hard, and so doesn't belong in the lead as an obvious restatement of the definition. --128.101.152.166 (talk) 17:00, 31 July 2015 (UTC)

I think it's important for the lead to link to the article on cosets, since the characterization of normal subgroups in terms of left and right cosets is often used. For example, Lang's "Algebra" introduces normal subgroups by looking at subgroups H such that $Hx=xH$ for all x. Hungerford's GTM "Algebra" has a clear statement of five equivalent conditions that define a normal subgroup, one of which is "every left coset of N in G is a right coset of N in G". Sławomir
Biały 17:39, 31 July 2015 (UTC)

Ok, fair enough. --128.101.152.166 (talk) 18:06, 31 July 2015 (UTC)

Simple question I'm overthinking

What formula would show me the percentage increase over time, ie there is a five percent increase in production every hour for a week? What would be the percentage increase in production at the end of the week? Beach drifter (talk) 18:11, 31 July 2015 (UTC)

I think you're talking about compounding here. For instance, if there is a 5% (0.05) increase in production every hour, starting at 100 arbitrary units, production will start increasing like this: 100, 105, 110.25, 115.7625, 121.550625. It doesn't just increase by 5 because each bit that's added increases the value of "5% of production". This type of "compound interest" is usually applied to interest on loans or deposits. The formula that I was taught, with "x" being the percentage change and "n" being the number of periods (e.g. every hour), was: (1 + x)ⁿ − 1. So with a 5% increase every hour, by the end of the week production will be (1 + 0.05)¹⁶⁸ = 3629.1 times its original value; this is a 362,810% increase in production rate. — Bilorv_(talk)^(c)(e) 18:37, 31 July 2015 (UTC)

• 362,813%. If you use that many digits, may as well make them all meaningful. —Tamfang (talk) 21:42, 1 August 2015 (UTC)

Agreed, but note that this means production stays at the initial rate until the hour ends, then jumps by 5%, stays at that rate for that hour, then jumps another 5%, etc. If instead of this step increase, it follows a steady increase, then a continuous compounding formula is needed, which gives a result over 4447 times the original. StuRat (talk) 21:07, 31 July 2015 (UTC)

Also, it is being assumed that production is continuous over the entire week (168 hours). --65.94.50.73 (talk) 22:52, 31 July 2015 (UTC)

No StuRat. (1 + x)ⁿ − 1 is already the correct formula for a steady increase. The formula for a step increase would be (1 + x)^⌊n⌋ − 1 (with a Floor function). To get 4447 times the original you need to assume a 5.127% increase in production every hour, which isn't what the example stated. Egnau (talk) 13:53, 1 August 2015 (UTC)

I read Bilorv's comments above to say that production for the first hour would be 100, for the second hour it would be 105, etc. This would mean it is moving up in steps. Otherwise, if production steadily increased at the rate of 5% per hour right from the start, then more than 100 units should be produced during the first hour (closer to 102.5 units). StuRat (talk) 16:35, 1 August 2015 (UTC)

There might be some ambiguity as to whether a production of "100" refers to an instantaneous rate of production, or to the number of units produced in 1 hour, but it doesn't matter as long as we stay consistent. In both cases, Bilorv's formula gives its increase over time assuming a steady increase. What you just did is inconsistent: you assumed that it refers to an instantaneous rate of production, you computed the number of units produced over an interval, and you reinterpreted that number as an instantaneous rate of production. Egnau (talk) 09:58, 2 August 2015 (UTC)

Egnau, you probably should read the article continuous compounding -- what StuRat is saying is correct and standard. --JBL (talk) 14:08, 2 August 2015 (UTC)

I know about continuous compounding and it's being misapplied here. Basically, in the steady increase case, the 5% figure already takes into account continuous compounding, so it's an error to apply the concept twice. To take a simpler example, if a lily pad doubles its size every day, then its size after 2 days is 4 times the original, irrespective of whether it grows steadily or in steps. The answer is not e² ≈ 7.389 times the original. Egnau (talk) 15:33, 2 August 2015 (UTC)

I'm with Egnau on this. If the increase is a continuous exponential, and we're told in the OP that the production increases by 5% every hour, then it should increase by 5% every hour, so the formula is $(1+0.05)^{n}=\exp(0.04879n)$. If, instead, we're told the instantaneous growth rate is 5%/hr, then it's $\exp(0.05n)$. Since the terms are very similar, there's an inherent ambiguity when using this sort of phrasing, so formulas should be used instead. -- Meni Rosenfeld (talk) 20:11, 2 August 2015 (UTC)

The term percent per hour is badly chosen because the increment in percent is not proportional to the time in hours. There does not seem to exist a generally accepted name for the unit of logarithm with base e^0.01. Using the name degree and the symbol ° you might say 5° per hour for the growth rate e^0.05n. Bo Jacoby (talk) 21:40, 2 August 2015 (UTC).

In mathematics, units are just numbers. % = 1/100, though mentioning percents usually clarifies that we are talking of a proportion of something else. The measures don't imply that growth is linear - it is a derivative, signifying the momentary rate regardless of what happens later. "Growth rate of 5%/hr" means that $\lim _{t\to 0}{\frac {f(n+t)-f(n)}{t}}={\frac {0.05}{\mathrm {hr} }}f(n)$ (the denominator is in units of time, so the result is also in units in time). From this we can deduce $f(n)=A\exp(0.05n/\mathrm {hr} )$ (or simply $f(n)=A\exp(0.05n)$ if we assume that hr=1). Whereas, "a five percent increase in production every hour" means that $f(n+\mathrm {hr} )=1.05f(n)$, so $f(n)=A\cdot 1.05^{n/\mathrm {hr} }$. -- Meni Rosenfeld (talk) 07:01, 3 August 2015 (UTC)

A constant growth rate of 5% per hour means that the variable y in x hours grows x·5%. So y = y₀·(1+0.05·x). This is linear growth.

A constant growth rate of 5° per hour means that the variable y in x hours grows x·5°. So y = y₀·e^0.05·x. This is exponential growth.

Bo Jacoby (talk) 21:05, 3 August 2015 (UTC).

The difference between the two cases should not be in the "5%/5°" part. They're both just numbers and they both mean 0.05. The difference should be in the "constant growth rate" part. We don't have very descriptive terminology unfortunately, but that's where we should be looking at. You can't have magic units that alter the meaning of the concepts around them. -- Meni Rosenfeld (talk) 21:55, 3 August 2015 (UTC)

Units do alter the meaning of the concepts around them. 5 hours is not equal to 5 miles. They're not just numbers. We agree that the lack of descriptive terminology is unfortunate. That's why I distinguish between percent and degree. Incrementing by 5% means multiplying by 1+5/100 = 1.05 while incrementing by 5° means multiplying by e^5/100 = 1.05127. -- Bo Jacoby (talk) 05:17, 4 August 2015 (UTC).

It's different in physics, physics have incommensurable quantities. But even in physics, the units don't alter the meaning of anything, they just can't be used with quantities of the wrong dimension. "The length of the rod is 5 hours" is gibberish; it is not a valid sentence that means something different from "the length of the rod is 5 miles". If you want to use hours you need to talk about something else, e.g. "the amount of time elapsed between events A and B is 5 hours".

Also, "being a number" is not the same as "being equal". We can argue about what "hour" really means, but we will all agree that 5 hours is equal to 300 minutes, and unequal to 5 minutes.

In the last sentence you've contradicted yourself. If incrementing by 5% means multiplying by 1.05, then incrementing by 5% every hour means multiplying by 1.05 every hour, so after x hours the amount is $1.05^{x}y_{0}$, and not $(1+0.05x)y_{0}$ as you previously wrote.

In any case, notation and terminology should make things clear. What you are proposing is IMO quite unlike anything else in mathematics, and as such would be the source of much confusion. You should instead be thinking of phrases like "incrementing the log by 0.05". -- Meni Rosenfeld (talk) 06:49, 4 August 2015 (UTC)

The meaning of incrementing the log by 0.05 depends on the base of the logarithm. If it is the base 10 logarithm you might say: incrementing by 0.05 decades. If it is the natural logarithm I would say: incrementing by 5°.

The contradiction is that incrementing twice by 5% is not the same thing as incrementing by 10%. This tells us that % is not what we want. But incrementing twice by 5° is the same thing as incrementing by 10°. Bo Jacoby (talk) 08:01, 4 August 2015 (UTC).